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Question Number 89596 by A8;15: last updated on 18/Apr/20

Commented by john santu last updated on 18/Apr/20

$$\sqrt{\frac{x}{1-x}}=t\Rightarrow x=\frac{t^2}{t^2+1}$$$$dx=\frac{2t}{{(t^2+1)}^2}dt]$$$$\int \frac{2t^2}{{(t^2+1)}^2}dt=$$$$\int t\left(\frac{d(t^2+1)}{{(t^2+1)}^2}\right)=\left[byparts\right]$$$$-\frac{t}{t^2+1}+\int \frac{dt}{t^2+1}=$$$$-\frac{t}{t^2+1}+{\tan }^{-1}\left(t\right)+c=$$$$-\frac{\sqrt{\frac{x}{1-x}}}{\frac{1}{1-x}}+{\tan }^{-1}\left(\sqrt{\frac{x}{1-x}}\right)+c$$$$-\sqrt{x-x^2}+{\tan }^{-1}\left(\sqrt{\frac{x}{1-x}}\right)+c$$

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